Question

If a=3.0mm, b=4.0 mm, Q1= 60 nC, Q2= 80 nC and q= 32nC in the figure below, what is the magnitude of the total electric force on q ?

The answer is 1.3, what is the full procedure?

Answer 1

magnitude of force exerted by Q1 on q, F1 = k*Q1*q/(a^2 +b^2)

= 9*10^9*60*10^-9*32*10^-9/(0.003^2 + 0.004^2)

= 0.6912 N

direction of F1, theta1 = tan^-1(a/b)

= tan^-1(3/4)

= 36.9 degrees below +x axis

magnitude of force exerted by Q2 on q, F2 = k*Q2*q/(a^2 +b^2)

= 9*10^9*80*10^-9*32*10^-9/(0.003^2 + 0.004^2)

= 0.9216 N

direction of F2, theta2 = tan^-1(a/b)

= tan^-1(3/4)

= 36.9 degrees above +x axis

so, the angle between F1 and F2, theta = theta1 + theta2

= 36.9 + 36.9

= 73.8 degrees

the magnitude of the total electric force on q,

|Fnet| = sqrt(F1^2 + F2^2 + 2*F1*F2*cos(theta))

= sqrt(0.6912^2 + 0.9216^2 + 2*0.6912*0.9216*cos(73.8))

= 1.30 N